Answer:
(D) Water from streams and rivers carry salt to the ocean
Explanation:
Hope this helps!
The radiation emitted is a beta particle with a -1 charge. <span>Beta particles have a </span><span>medium penetrating power. An emission of beta particles requires shielding because of the hazards it pose to humans. Thus, one characteristic of this radiation is that some shielding is required.</span>
2 Al + 3 H₂SO₄ = Al₂(SO₄)₃ + 3 H₂
2 moles Al -------- 3 moles H₂SO₄
4.4 moles Al ------ ?
moles H₂SO₄ = 4.4 x 3 / 2
moles H₂SO₄ = 13.2 / 2
= 6.6 moles of H₂SO₄
hope this helps!
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
The molar concentration of the KI_3 solution is 0.0833 mol/L.
<em>Step 1</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-) = 25.00 mL S_2O_3^(2-) ×[0.200 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)] = 5.000 mmol S_2O_3^(2-)
<em>Step 2</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 5.000 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 2.500 mmol I_3^(-)
<em>Step 3</em>. Calculate the <em>molar concentration of the I_3^(-)</em>
<em>c</em> = "moles"/"litres" = 2.500 mmol/30.00 mL = 0.083 33 mol/L
